Question

The rate constant for a first order reaction is 60 s-1. How much time will it
take to reduce 1g of the reactant to 0.0625 g?

Answer 1

a1 = 0.0625 gm
a2 = 1 gm

r = 60 S-1

We know for first order reaction

d(a)/ d(t) = r
Thus
  d(t) = d(a)/r
       = a2-a1 / r
        = 1-0.0625 / 60
         = 0.015 Sec

Thus time is 0.015 Sec

Answer 2

Answer: 0.046 sec

Explanation: For a first order reaction, the rate law is:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant=$60sec^-1$

t = time taken for decay process= ?

a = initial amount of the reactant  = 1 g

a - x = amount left after decay process= 0.0625 g

Putting values in above equation, we get:

$t=\frac{2.303}{60}\log\frac{1}{0.0625g}=0.046sec$